1. The Minimum-Area Spanning Tree Problem
何宗融 江宗憲 黃安達 沈孟俞 蔡予欣 蔡鎮宇

2. Introduction Minimum-Area Spanning Tree(MAST):
Given a set P of n points in the plane, find a spanning tree of P of minimum area.
Area:
union of the n-1 disks whose diameters are edges in T
Main result:
minimum spanning tree of P is a constant-factor approximation for MAST
= Area

3. Extension problems Power assignment problem
Before Minimum-Area Range Assignment(MARA)
P: a set of n points( transmitters-receivers)
Goal:
The resulting directed communication graph is strongly connected pj
Radius of Dp:
the transmission range assign to Pi Dp pk pi

4. Extension problems Power assignment problem Goal:
The total power consumption is minimal
= Power consumption of power assignment problem
Result: NP-hard, 2-approximation based on MST + + + +

5. Extension problems Minimum-Area Range Assignment(MARA) Goal:
power assignment problem in radio networks
Goal:
Minimize the union of the disks Dp1,..Dpn(total coverage area)
Prevent from foreign receiver
= Power consumption of power assignment problem + + + +
coverage area of MARA =

6. Extension problems Minimum-Area Connected Disk Graph(MACDG) Goal:
the resulting disk graph is connected.
Minimize the union of the disks Dp1,..Dpn
Dpj pj pk pi
Dpi
Dpk

7. Extension problems Minimum area tour (MAT) Goal:
Variant of traveling salesman problem
Goal:
Minimize a tour of P of minimum area
constant-factor approximation based on relaxed triangle inequality
Dpj pj
Dpk pk pi
Dpi

8. MST v.s. MAST MST: Minimum Spanning Tree
MAST: Minimum Area Spanning Tree
We’ll prove that MST is a c-approximation for MAST

9. MST v.s. MAST (Cont.)

10. Some Definitions Let T be any spanning tree of P For an edge e in T
D(e) is the disk whose diameter is e
D(T) = {D(e) | e is an edge in T}
UT = Ue∈TD(e)
σT = ∑e∈Tarea(D(e))
We’ll prove that MST is a c-approximation for MAST
area(UMST) = O(area(UOPT))

11. Claim 1 Let MSTp be a MST for P ∪ {p}
There is no edge (a, b) in MSTp, such that (a, b) is not in MST and both a and b are points of P

12. Claim 1 Proof Assume there is an edge (a, b) in MSTp but not in MST B

13. Claim 1 Proof (Cont.) Consider the path in MST between a and b
At least one of the edges along this path is not in MSTp (edge e) e B B A A
MSTP
MST

14. Claim 1 Proof (Cont.) |e| < |(a, b)|
We can replace (a, b) in MSTp by e, without increasing the total weight e e B B A A
MSTP
MST

15. An Corollary of Claim 1
If e is an edge in MSTp but not in MST, then one of e’s endpoints is p

16. Lemma 1 σMST ≤ 5 area(UMST)
At first, we need to prove that p belongs at most 5 of the disks in D(MST)

17. Lemma 1 Proof Let D(q1, q2) be a disk in D(MST)
If p ∈ D(q1, q2), the edge(q1, q2) is not in MSTp
If it is, we can replace (q1, q2) by (p, q1) or (p, q2) to decrease total weight of MSTp q1 q2 q1 q2 p p

18. Lemma 1 Proof (Cont.) By the corollary of Claim 1
If e is an edge in MSTp but not in MST, then one of e’s endpoints is p
Each disk D ∈ D(MST) such that p ∈ D, induces a distinct a distinct edge in MSTp
The degree of p is at most 6
This is true for any vertex of any Euclidean MST

19. Lemma 1 Proof (Cont.) So there can be at most 5 disks covering p
Then we prove that σMST ≤ 5 area(UMST)

20. ST Construction
Let OPT be an optimal spanning tree of P, i.e., a solution to MAST.
We use OPT to construct another spanning tree ST of P

21. ST Construction (Cont.)
Initially ST is empty
In the i’th iteration, let ei be the longest edge in OPT and there is no path in ST between its endpoints
Draw two concentric circles Ci and Ci3 around the mid point of ei
The diameter of Ci is |ei|
The diameter of Ci3 is 3|ei|

22. ST Construction (Cont.)
Apply Kruskal’s MST algorithm with the modification to the points of P lying in Ci3
The edge can’t be already in ST
The edge can’t create cycle in ST
Si is the edge set return by Kruskal algorithm in i’th iteration, and we add Si to ST

23. ST Construction Examples

24. ST Construction Examples

25. ST Construction Examples

26. Claim 2
For each i, is a subset of the edge set of the minimum spanning tree that is obtained by applying Kruskal’s algorithm, without the modification above, to the points in

27. Claim 3: ST is a spanning tree of P
Proof - there are no cycles in ST. - ST is connected, since otherwise there still exists an edge in opt that forces another iteration of the construction algorithm.

28. Claim 4

29. Claim 5 = ( C is a subset of D(OPT) ) (by lemma 1) (by claim 2)

30. Theorem 1
MST is a constant-factor approximation for MAST
(by claim 5)

31. Constant-Factor Approximation for Minimum-Area Range Assignment
Let pi  P and ri is the length of the longest edge in MST that is connected to pi.
RA = { Dp1,…,Dpn}, where Dpi is the disk of radius ri, centered at pi.
Let OPTR denote an optimal range assignment.

32. MARA problem Problem definition
The corresponding(directed) communication graph is strongly connected.
The area of the union of the disks in RA is bounded by some constant times the area of the union of the transmission disks in an optimal range assignment.

33. Claim 6: area(URA) ≤ 9∙area(UMST)

34. Claim 6: area(URA) ≤ 9∙area(UMST)
Proof
The area(Dpi(pi,pj)) ≤ area(D3(pi,pj)) = 9∙areaD(pi,pj)
area(URA) ≤ 9area(UMST)
Area(UMST)
Area(URA)

35. Proof: area(URA) ≤ c’∙area(UOPTR)
Theorem 2. RA is a constant-factor approximation for MARA, i.e., area(URA) ≤ c’∙area(UOPTR), for some constant c’
Proof: area(URA) ≤ c’∙area(UOPTR)
We construct a spanning tree T of P as following,
For each point q  P, q ≠p, compute a shortest directed path from q to p, and add the path to T.
Make all edges in T undirected.
Hence, UT  UOPTR
area(URA) ≤1 9area(UMST) ≤2 9c∙area(UOPT) ≤3 9c∙area(UT) ≤4 9c∙area(UOPTR)

36. Inequality 1 Inequality 2 Inequality 3 Inequality 4
area(URA) ≤1 9area(UMST) ≤2 9c∙area(UOPT) ≤3 9c∙area(UT) ≤4 9c∙area(UOPTR)
Inequality 1
According to claim 6 (area(URA) ≤ 9∙area(UMST))
Inequality 2
Follows from Theorem 1 (area(UMST)≤c∙area(UOPT))
Inequality 3
From the definition of OPT
Inequality 4
Show above

37. A Constant-Factor Approximation for MACDG

38. MACDG Minimum-Area Connected Disk Graph (MACDG) problem Pl D(pk, pl)
Pi Pj Pl Pk
D(pi, pj)
D(pi, pk)
D(pk, pl)

39. MACDG: Goal and Define
Goal: Decrease the coverage of overlapping from MARA.
Define:
DG = {Dp1,..,Dpng}, where Dpi is the disk of radius ri/2 centered at pi
OPTD denote an optimal assignment of radii, i.e., a solution to MACDG.
Pi Pj Pk
D(pi, pj)
D(pi, pk) Pl
D(pk, pl)

40. MACDG: A Constant-Factor Approximation
Requirements:
(i) DG is connected
(ii) the area of the union of the disks in DG is bounded by some constant times the area of the union of the disks in an optimal assignment of radii
The 1st requirement above clearly holds, since each edge in MST is also an edge in DG.
And the 2nd requirement…(Theorem 3)

41. MACDG: Theorem 3
DG is a constant-factor approximation for MACDG, area(UDG) ≤ c’’∙ area(UOPTD), for some constant c’’
Proof:
(Claim 6) It is very similar to the proof of MARA. Since UDG  URA, area(UDG) ≤ 9 ∙ area(UMST)
(Theorem 1) area(UMST) ≤ c ∙ area(UOPTD) area(UDG) ≤ 9∙area(UMST) ≤ 9c * area(UOPTD) = c’’∙ area(UOPTD)

42. Constant-Factor Approximation for MATe G2

43. Relaxed Triangle Inequality

44. Constant-factor Approximation Algorithms for the TSP
For distance functions: d(u,v) ≤ τ∙(|uw|2+|wv|2)
Andreae and Bandelt: (3τ2/2+τ/2)-approximation
Andrea: (τ2+τ)-approximation
Bender and Chekuri: 4τ-approximation
This implies there is a 6-approximation for our case

45. Constant-Factor Approximation for MAT
Andreae and Bandelt computed a tour T in G2 such that w(T) ≤ c∙w(MSTG2)
T is a constant-factor approximation for the Minimum Area Tour (MAT) problem

46. Notations D(e) denotes the disk whose diameter is e
D(T) = { D(e) | e is an edge in T }
T = eT D(e)
σT = ΣeT area(D(e))
MST is the minimal spanning tree of P
OPTT is an optimal tour, i.e., a solution to MAT
OPTS is a solution to the MAST problem
Clearly, area(OPTT) ≥ area(OPTS)

47. Proof
but (by Lemma 1) By the main result of section 2 (MAST)

48. Theorem 4
T is a constant-factor approximation for MAT, i.e., area(T) ≤ ĉ∙area(OPTT)